I would like to output list of strings value to ostream.
I can declare and implement overloading function for this:
ostream& operator<< (ostream &out, const list<string> &in);
... and then write like
cout << value;
... but there are at least two possible ways to dump list of strings: one string per line or all strings in one line separated by spaces (or maybe other separator).
Is it possible to change dumping function behavior through stream controlling?
I would like to write something like:
list<string> lst;
...
cout << print_as_multiline() << lst;
... and:
list<string> lst;
...
cout << print_as_one_line() << set_separator (", ") << lst;
I have an idea… could you test it?
In your operator << implementation file Create a global,static?? variable string Separator= def sep (", ").
Use it in your <<, (and before return reset to def sep?- your decision)
Create a class set_separator, with the implementation of the constructor in the same file as your <<. In this constructor set Separator to the argument.
Define a new operator << for class set_separator, with do nothing, just return the stream.
Similar with multiline output.
There are several options for that:
- Write multiple functions or a function taking a separator as parameter.
- Attach the separator to the stream. You can use the iostream's xalloc() function to allocate a custom property slot (it should be allocated only once and then applies to all streams). You can then use the streams' iword() and pword() memberfunctions to access the attached info. There is also an event that you can hook into, so a custom function is called when e.g. the stream is destroyed. Use that in order to release dynamically allocated content, in case that is necessary.
- Create a so-called facet that takes care of the list formatting. This facet is attached to the stream's locale.
My advise: Take the first option, it's the least complicated and most straightforward one. If you need to decide the formatting in one place but actually use it in a completely different place, use the second one. Read Langer&Kreft C++ IOStreams and Locales before you consider the third option. ;)
Related
I've been working on a personal dictionary application which can help you remembering words you learnt. It is operated via the CLI (just don't question this, it's kinda just a test and I got a weird passion for CLI apps). So, of course I am using ostreams for writing information on the CLI. I am used to write operator<< overloads (for ostreams) for every class so that I can build up a multi-level output system (basically every object can "speak" for itself).
In order to persist a dictionary object, I wanted to use ofstream and write a file with it. Naturally, I wrote operator<< overloads also for ofstream and in the same "layered" structure.
As a result, I have now two operator<< overloads in every class, like in "Dictionary":
ostream& operator<<(ostream&, const Dictionary&);
ofstream& operator<<(ofstream&, const Dictionary&);
(this is just the declaration in the header file)
Notice that it is very important that these both overloads do different things. I don't want to have some weird persistence-oriented special-format text on the CLI and also not user-friendly plain text in my file.
The problem is that, because of the inheritance structure of ostream and ofstream, ofstream is sometimes implicitely converted to ostream. And when this happens in the middle of my stack full of file output operations, the program suddenly jumps into the wrong overload and prints plain text in the file.
My question is simply: Is there a way to avoid or revert these unwanted implicit conversions in order to let my program jump into the right overloads? Or is there any other good way to fix this problem?
EDIT 1:
Someone pointed out in the comments that this is not an implicit converison. ofstream is sometimes "seen" as its base class ostream. The problem is that at some point the object "forgets" that it is an ofstream and loses all file-related information. From there on it is only an ostream and that's what I meant with the "conversion".
EDIT 2:
The exact point in the program where the "unwanted conversion" happens can be found here:
ofstream& operator<<(ofstream& of, const Section& s) {
return s.print_ofstream(of);
}
So this operator overoad calls "print_ofstream":
ofstream& Section::print_ofstream(ofstream& of) const {
of << "sec" << Util::ID_TO_STRING(section_id) << ":\n";
for (pair<Wordlist, Wordlist> pwl : translations) {
of << '{' << pwl.first << '=' << pwl.second << "}\n";
}
of << "#\n";
return of;
}
Note that "pwl" is a pair of two Wordlists, therefore pwl.first / pwl.second is a Wordlist. So, normally the line of << '{' << pwl.first << '=' << pwl.second << "}\n"; should call the ofstream operator<< overload in Wordlist. But it doesn't. Instead, the other overload method is called:
ostream& operator<<(ostream& o, const Wordlist& wl) {
return wl.print_ostream(o);
}
You have overloaded only the specific operator<< needed for streaming Dictionary, Section, Wordlist, etc objects to a std::ofstream, but std::ofstream inherits MANY other operator<<s from std::ostream, and those operators all take ostream& as input and return ostream& as output. So, for example, of << "sec" will return ostream& even though of is a std::ofstream, and then that ostream& is used for subsequent << calls until ; is reached. Those are the "implicit conversions" you are experiencing.
The real question is, WHY do you want operator<< to output different data depending on the type of std::ostream being written to? That goes against C++'s streaming model. If you really want that, you would have to change print_ofstream(ofstream&) to print_ostream(ostream&) and then dynamically detect the actual std::ostream derived type using dynamic_cast. Same with Wordlist, and any other classes that need it.
A simpler and safer option would be to just store a flag inside of your classes to control how their data should be output, regardless of the type of std::ostream being used. Then you can set that flag as needed. Maybe even define some helper I/O manipulators to set those flags while making << calls.
I am using ifstream and and ostream to serialize my data but I am surprised to discover the `<<' operator can't seperate two adjacent strings and seperating them would be quite complicated.
class Name
{
string first_name;
string last name;
friend std::ostream& operator<< (std::ostream& os, const Name& _name)
{
os << _name.first_name << _name.last_name;
return os;
}
friend std::istream& operator>> (std::istream& is, Name& _name)
{
is >> _name.first_name >> _name.last_name;
return is;
}
This doesn't work because << and >> doesn't write null terminator characters and ifstream reads the whole string in variable (first_name) which is kinda disappointing. How can I store the two strings separately so I can read them separately as well? I don't understand what is the motivation of << concatenating all the strings in ostream so we can't read them back seperatly!?
I don't understand what is the motivation of << concatenating all the strings in ostream so we can't read them back seperatly!?
This assumes that the only reason to write them separately is to read them as individual strings. Consider the case where someone has a pair of strings that they want to write to a stream without separators. Or a string followed by a float that they don't want separators for.
If ostreams automatically inserted separators for every << output, then it would be much harder for someone to write text without separators. They'd have to manually concatenate these strings and/or values into a single string, then output that.
And what would they use for this concatenation? They can't use ostringstream like you normally might, because it uses the same facilities as ofstream. So every << would put a separator character in the stream.
In short, the IO streams API writes what you told it to write, not what you may or may not "want" to write. It's not a serialization API; C++ isn't C# or Java. If you want serious serialization features, use Boost.Serialization.
Often times you want to concatenate strings with ostream (commonly stringstream). If you specifically don't want them concatenated it's easy enough to do:
os << _name.first_name << '\n' << _name.last_name;
ifstream and ofstream basically are streams, so they have nothing to indicate limit of data in them. Think about them as a river, all data can read from or write to them. This is true nature of files, so if you need them for serialization you must implement your serialization mechanism or use a library that designed for this purpose like boost::serialization. In C++ every thing implemented as is, and because of this you can gain maximum performance!! :)
I am still confused about the difference between ostream& write ( const char* s , streamsize n ) in c++ and cout in c++
The first function writes the block of data pointed by s, with a size of n characters, into the output buffer. The characters are written sequentially until n have been written.
whereas cout is an object of class ostream that represents the standard output stream. It corresponds to the cstdio stream stdout.
Can anyone clearly bring out the differences between the two functions.
ostream& write ( const char* s , streamsize n );
Is an Unformatted output function and what is written is not necessarily a c-string, therefore any null-character found in the array s is copied to the destination and does not end the writing process.
cout is an object of class ostream that represents the standard output stream.
It can write characters either as formatted data using for example the insertion operator ostream::operator<< or as Unformatted data using the write member function.
You are asking what is the difference between a class member function and an instance of the class? cout is an ostream and has a write() method.
As to the difference between cout << "Some string" and cout.write("Some string", 11): It does the same, << might be a tiny bit slower since write() can be optimized as it knows the length of the string in advance. On the other hand, << looks nice and can be used with many types, such as numbers. You can write cout << 5;, but not cout.write(5).
cout is not a function. Like you said, it is an object of class ostream. And as an object of that class, it possesses the write function, which can be called like this:
cout.write(source,size);
"In binary files, to input and output data with the extraction and insertion operators (<< and >>) and functions like getline is not efficient, since we do not need to format any data, and data may not use the separation codes used by text files to separate elements (like space, newline, etc...).
File streams include two member functions specifically designed to input and output binary data sequentially: write and read. The first one (write) is a member function of ostream inherited by ofstream. And read is a member function of istream that is inherited by ifstream. Objects of class fstream have both members. Their prototypes are:
write ( memory_block, size );
read ( memory_block, size );
"
from: http://www.cplusplus.com/doc/tutorial/files/
There is no function ostream& write ( const char* s , streamsize n ). Perhaps you are referring to the member function ostream& ostream::write ( const char* s , streamsize n )?
The .write() function is called raw (or unformatted) output. It simply outputs a series of bytes into the stream.
The global variable cout is one instance of class ofstream and has the .write() method. However, cout is typically used for formatted output, such as:
string username = "Poulami";
cout << "Username: '" << username << "'." << endl;
Many different types have the ostream& operator<<(ostream& stream, const UserDefinedType& data), which can be overloaded to enrich ofstream's vocabulary.
Oh boy! A chance to smash up a question.
From your question I feel you are some Java or Python programmer and definitely not a begginner.
You dont understand that C++ is probably the only language that allows programmers to implement primitive built in operators as class members and as part of the general interface.
In Java you could never go
class Money
{
int operator + (int cash) { return this.cash + cash; }
void operator << () { System.out.println(cash); }
int cash;
}
public class Main_
{
public static void Main(String [] args)
{
Money cashOnHand;
System << cashOnHand;
}
}
But cpp allows this with great effect. class std::ostream implements the stream operators but also implements a regular write function which does raw binary operations.
I agreed with Alok Save！A litte before, I searched the problem and read the answer carefully.
Maybe in other word, cout is an object of ostream, but write is just a function provided. So cout have twe ways to used by coders: one is as a member function, another is used by operator(<<).
I have an object that can be printed to the console with std::cout << obj, but I can't get a std::string out of it, because it doesn't seem to implement something like a .string() method. I thought I might be able to use that overloaded operator to just get string representations of everything instead of having to implement a function to do it myself every time I need it, though having found nothing on the subject makes me think this isn't possible.
Use a std::ostringstream. It is a C++ stream implementation which writes to a string.
You can use a std::ostringstream.
std::ostringstream os;
os << obj;
std::string result = os.str();
There are different ways of doing it, you can manually implement it in terms of std::ostringstream, or you can use a prepacked version of it in boost::lexical_cast. For more complex operations, you can implement a in-place string builder like the one I provided as an answer here (this solves a more complex problem of building generic strings, but if you want to check it is a simple generic solution).
It seems that the linked question has been removed from StackOverflow, so I will provide the basic skeleton. The first think is to consider what we want to use with the in-place string builder, which basically is avoiding the need to use create unnecessary objects:
void f( std::string const & x );
f( make_string() << "Hello " << name << ", your are " << age << " years old." );
For that to work, make_string() must provide an object that is able to take advantage of the already existing operator<< for the different types. And the whole expression must be convertible to std::string. The basic implementation is rather simple:
class make_string {
std::ostringstream buffer;
public:
template <typename T>
make_string& operator<<( T const & obj ) {
buffer << obj;
return *this;
}
operator std::string() const {
return buffer.str();
}
};
This takes care of most of the implementation with the very least amount of code. It has some shortcomings, for example it does not take manipulators (make_string() << std::hex << 30), for that you have to provide extra overloads that take the manipulators (function pointers). There are other small issues with this implementation, most of which can be overcome by adding extra overloads, but the basic implementation above is enough for most regular cases.
i have an overloaded operator << trying to make it work like this
mystream<<hex<<10;
i have overloaded method
mytream& operator<<(ios_base& (*m) ios_base&)
This gets called whenever hex is encountered cause the parameter passed in the method is a function pointer of type same as hex or like some other output manipulators like dec, oct.
i have two problems
1) how do i retrieve the parameter the hex would be operating on, in this example 10
2) how do i know that the << operator is being called for hex and not other manipulator function like oct and dec
Thanks
1) hex is not operating on the parameter 10. << operators associate left-to-right, which means your code is the same as:
(mystream<<hex)<<10;
So your overload has to return an object which, when 10 is shifted into it, prints in hex (or if not prints, writes data somewhere). As everyone says, this is done by saving flags in the stream object itself, then returning *this. The reason flags are used is precisely because the "10" is not available yet, since the second << has not been evaluated yet. The first << operator call cannot print anything - it just has to get ready for when the second one is called.
2) hex is a function. It can be compared with other functions:
ostream &operator<<(ostream &s, ios_base& (*m)(ios_base &)) {
if (m == hex) {
} else if (m == oct) {
} else if (m == dec) {
}
}
Except you don't normally want to do that, you want the default behaviour, which is something like:
ostream &operator<<(ostream &s, ios_base& (*m)(ios_base &)) {
return m(s);
}
(I may be wrong on that, I've never looked at the implementation, but the general idea is that the operator calls the manipulator function, and the manipulator (the clue's in the name) manipulates the stream).
std::hex sets the std::ios::hex format flag on its parameter. Then in your operator<<(int) override, if you have one, check the format flags by calling flags().
3) Manipulators which take paramers are functions too, but their return types are unspecified, meaning it's up to the implementation. Looking at my gcc iomanip header, setw returns _Setw, setprecision returns _Setprecision, and so on. The Apache library does it differently, more like the no-args manipulators. The only thing you can portably do with parameterized manipulators is apply them to an iostream with operator<<, they have no defined member functions or operators of their own.
So just like hex, to handle setw you should inherit from std::ios_base, rely on the operator<< implementation provided by your library, then when you come to format your data, examine your own width, precision, etc, using the width(), precision(), etc, functions on ios_base.
That said, if for some bizarre reason you needed to intercept the standard operator<< for these manipulators, you could probably bodge something together, along these lines:
template <typename SManip>
mystream &operator<<(mystream &s, SManip m) {
stringstream ss;
// set the state of ss to match that of s
ss.width(s.width());
ss.precision(s.precision());
// etc
ss << m;
// set the state of s to match that of ss
s.width(ss.width());
s.precision(ss.precision());
// etc
return s;
}
I do consider this a bodge, though. You're not really supposed to interfere with stream manipulators, just let your base class do the work and look up the results.
When operator<< gets called with hex or oct or dec, set a flag in your mystream object. When operator<< is called with a number, check to see if any of these flags are set. If so, convert the number to hex/octal/decimal and display it.
In answer to your second question, the parameter m is a pointer to the manipulator function. You can check that it's not null, then call that function, passing *this. hex() is as simple as setting a flag in the passed stream object, as Zifre suggested. Then when processing the integer, check if the flag in the stream object is set, and output accordingly.
This is how the standard library implements its manipulator functions.
In your example, hex operates on (changes the state of) the stream, not the following parameters. hex has no notion of, or any relation to other << calls.
Looking at how other io manipulators are implemented would go a long way to clearing things up.
You should be manipulating ios_base::flags
http://www.cplusplus.com/reference/iostream/ios_base/flags/
which is what the standard hex does.